# Photon sphere vs horizon as a null suface

## Main Question or Discussion Point

Hi ,
There's something I don't get regarding the orbits of photons in Schwarzschild geometry.

As well known, by solving geodesics equation for null rays, you get that photons can be in a (unstable) circular orbit at r=3 M. However, if you look at the causal diagram for Schwarschild geometry (in Schwarzschild coordinates, say), then you see that the light cones are still open.

So, according to this diagram, a photon should go left (towards decreasing r) or right (increasing r), but not straight at r=cst!! Is the solution somewhere hidden in the two other coordinates theta and phi?

I have another, but related question. The light cone gets degenerate with no extension at r=2M, as you see again on this causal diagram. In other words, the horizon is a null surface, which is a general property anyway. But if the horizon is a null surface, why don't photons actually move along this null surface? Instead, they just cross inwards.

To summarize, I would have expected the photon sphere to be precisely the horizon, since for me I have this (wrong) idea in head : null surface <-> photon sphere

Where is the mistake?

Thanks a lot

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Bill_K
At the horizon r = 2M, the light rays (not "photons") that generate the null surface travel outward. Remaining at constant r = 2M of course.

r = 3M is not a null surface. The light rays that go sideways travel in circular orbits.

Last edited:
DrGreg
Gold Member
So, according to this diagram, a photon should go left (towards decreasing r) or right (increasing r), but not straight at r=cst!! Is the solution somewhere hidden in the two other coordinates theta and phi?
Yes. The diagram shows only radial motion (constant $\theta$ and $\phi$) and so can't show orbits. The light cone is a 4-D cone and the diagram you refer to (I assume) shows only 2 dimensions.

bcrowell
Staff Emeritus
Gold Member
I have another, but related question. The light cone gets degenerate with no extension at r=2M, as you see again on this causal diagram. In other words, the horizon is a null surface, which is a general property anyway. But if the horizon is a null surface, why don't photons actually move along this null surface? Instead, they just cross inwards.
A light ray directed inward from the horizon travels inward. A light ray directed outward from the horizon maintains constant r=2M, so on a Penrose diagram it just travels along that diagonal line until it gets to $\mathscr{I}^+$.

A light ray directed outward from the horizon maintains constant r=2M, so on a Penrose diagram it just travels along that diagonal line until it gets to I+
Hi,
Thanks for the replies. Now I understand better.

-For the photon sphere, a light ray can travel inward or outward at r=3M, but it might also somehow balance gravity with some radial speed + an orthoradial speed (roughly speaking, it's just a way to visualize what's going on; I think the anology with an accoustic black hole helps here, where I guess you can have a circular sonic wave orbiting around the sonic horizon (?))

-Regarding the light cones at r=2M. I think I was fooled by the use of Schwarzschild coordinates. In these coordinates, the light cone schrinks and has no extension anymore at r=2M. So, naively, you think that light rays must stay at r=2M. However in advanced Eddigthon-Finkelstein coordinates, you rather see that the light cone is not degenerate, and that, as you say, the light ray outward stays at r=2M, all the others fall towards the singularity. I have to see this as well on the Penrose diagram.

My confusion came from the fact that if you stay at r=2M, then you are in a circular orbit. But it is wrong. You can be static as well!

So, if I understood correctly, there are null geodesics that stay forever at some point $theta_0$, $\phi_0$, $r=2M$, $\forall t$. Is that right?